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lang-js
Arraydifference is a so calledset operation, because property lookup is the very own job ofSets, which are orders of magnitude faster thenindexOf/includes. Simply put, your solution is very inefficient and rather slow.Set, values have to be unique, no?[1,2,3] [2,3,5]given that the numbers are unique but if you had say[1,1,2,3] [1,2,3,5]and expected[1]you couldn't useSet. Your solution wouldn't work either though :-/ I ended up creating this function because I couldn't figure out a satisfactory way to do it more succinctly. If you have any ideas on how to do that, I'd love to know!Array.includes()ES7 feature instead of ES6? (1) (2) — and to continue, with ES6 you could useArray.some()e.g.let intersection = aArray.filter(a => bArray.some(b => a === b)), no?includes, which obviously gave massive speed increases (improved the complexity fromO(n^2)toO(n log2 n)). Using her little function instead of binary search came out 2-3x times faster, especially when the arrays were almost the same. 10x+ when using crazy arrays (100,000,000 elements)...